HELP ME HELP ME GEOMETRY PROBLM!!!
posted by SUPER HARD .
In triangle ABC, m<A = 90, m<B=75, and BC = \sqrt{3}. What is the area of ABC? Express your answer as a common fraction.
HELP THIS IS SO HARD HHELP MEE!!! EXplain too please!

You have a rightangled triangle so we can use the formula
(1/2) base x height
Since you want the answer as a common fraction, I will assume that you need "exact" values.
you state that BC = \sqrt{3}
I will assume that is meant to say √3
so cos75 =base/√3
base = √3cos75
and sin75 = height/√3
height = √3sin75
area = (1/2)(√3sin75)(√3cos75)
= (3/2)sin75cos75)
(recall sin 2A = 2sinAcosA)
= (3/2)(2sin75cos75)/2
= (3/4) sin 150
sin 150 = sin30  one of the basic identities
area = (1/3) sin150
= (3/4)sin30
= (3/4)(1/2) = 3/8 
thanks alot

THANK YOU !!!!!!!

You don't even need trig for this. Reflect B, draw and altitude from B to B'C and solve